and the maximum value of the cosine is ±1, Using Eq. 1, when the empty car oscillates onso the springs, the frequency will be amax = A ω 2 . r 1 k f=Therefore, from Newton’s 2nd law above, or 2 πrM 1 9mg µs m g = m Amax ω 2 , so = 2π M x µs g s Amax = 2 1 (747 kg) (9.8 m/s2 ) ω = (0.324)(9.8 m/s2 ) 2 π (2398 kg) (0.0089 m) = (8.16814 rad/s)2 = 2.94764 Hz . = 4.7591 cm . Question 8, chap 15, sect 1. Note: Neither given mass is required to part 1 of 3 10 pointswork this problem. In an experiment conducted on the space Question 7, chap 15, sect 1. shuttle (i.e., in free fall), a horizontal rod of part 1 of 1 10 points mass 52 kg and length 45 m is pivoted about a point 6.8 m from one end, while the opposite When nine people whose average mass is end is attached to a spring of force constant83 kg sit down in a car, they find that the car 98 N/m and negligible mass.drops 0.89 cm lower on its springs. Then theyget out of the car and bounce it up and down. The acceleration of gravity is 9.8 m/s2 . 45 m What is the frequency of the car’s vibration 52 k g 98 N/m

Let : m = 9.2 kg , ℓ = 10.9 m , d = 4 m, ℓ − d = 6.9 m , θ = 46 ◦ , and g = 9.8 m/s2 . As shown in the figure, the motion can bebroken into two parts: that before the stringcontacts the peg and that after it makes con-tact. Its motion on the right half of thediagram is that of a pendulum with length10.9 m and period T1 ; its motion on the left